Cơ học lượng tử 2.1

Energies of a particle in a finite spherical well Read Chapter 13... Energies of a particle in a finite spherical square well... Because E is intrinsically negative, we can write: E = -

I.  What to recall about motion in a central potential

II.  Example and solution of the radial equation for a particle trapped within radius “a” III.  The spherical square well

(Re-)Read Chapter 12 Section 12.3 and 12.4

1


I What to recall about motion in a central potential

2 [ V (r) − E ] = ( + 1) "Form 1" of the radial equation

the angular momentum operator

3


You can get an alternative completely equivalent form of this equation if you derive

⎥u = Eu "Form 2" of the radial equation

*Choose either Form 1 or Form 2 depending upon what V

is pick whichever gives and easier equation to solve

*Remember the boundary conditions on R:

rR(r → ∞) → 0 BC1

rR(r → 0) → 0 BC2

Procedure for finding the total Ψ(r,t) for a system in a central potential:

(i) Get V(r)

(ii) Plug it into the radial equation (either Form 1 or Form 2),

solve for R and the energies Ei

(iii) Multiply that R by Ym( θ,ϕ) and e−iE i t

 to get Ψ(r,t)=RYe−iE i t



4


II Example-Solution of the radial equation for a particle trapped within radius "a" V= 0 for r < a

∞ for r > a

⎧

⎨

⎩

This is also called a "spherical box"

Recall the radial equation in Form 1 (without the substitution u=rR):

Since V=∞ for r > a, the wave function cannot have any portion beyond r > a

So just solve the equation for r < a.

Define ρ ≡ kr, so 1

ρ Then d

"spherical Bessel function" "ordinary Bessel function

of half-odd integer order"

Normalization not yet specified

Spherical Neuman function, Irregular @ r=0, so get D=0

I.  Particle in 3-D spherical well (continued)

II.  Energies of a particle in a finite spherical well

Read Chapter 13

Now apply the BC: the wave function R must = 0 @ r = a

j(ka) = 0

whenever a Bessel function = 0 its argument (here: (ka)) is called

a "zero" of the spherical Bessel function

we have to specify n too,

so R ≡ Rn

In spectroscopy

the rows are

labelled by:

Recipe to find an allowed energy:

I Energies of a particle in a finite spherical square well

This actuall looks very much like the potential between 2 nucleons in the nucleus

Solve this analogously to 1-D square well procedure

(i) Define regions I and II

(ii) Plug in V into radial equation in each region

Pick Form 2, so solve for u=rR

(iii) Apply BC @ r = 0, a, ∞

This leads to quantized E

(we'll stop here)

(iv) Normalize if you want the exact form of Ψ

Case I: Find allowed energies of a particle in an  = 0 state of a finite spherical square well

(i) define regions I and II

Notice that the particle in the well will have E < 0 just because of the way we defined the potential

*This is just a convention (i.e a choice of the origin for the V scale), but since it is common we will use it Notice this is a different convention that the one we used for the 1-D square well

r

I II

0

0+V0

Because E is intrinsically negative, we can write:

E = -|E| when we wish

Recall Form 2 of the radial equation:

This is exactly the same form as for the 1-D square well,

so the wavefunctions "in" will have the same form as

the "Ψ's" for the 1-D square well

Divide equation 1equation 2 :

as the Class 2 (or odd) solutions to the 1-D square well.

Multiply both side by a:

k1a cot k1a = -k2a

-cotk1a = k2a

k1a

Intersection points are solution to the equation:

(use the identity that -cotx = +tan(π

y

0 π/2 π 3π/2 2π 5π/2 3π 7π/2 4π

LHS(-cot)

Call these yi Example:

Then plug in the definition of y:

yi = a

 2µ(V0− | E i |)

Plug in these values of a, u, V0, , to solve for Ei

y1 y2 y3

Alternatively if you measure the Ei (for example the bound state energies of a deuteron),

you can work backwards to find V0a2.

Recall Case 1 was for  = 0 only Now,

Case 2: Find the allowed energies for finite spherical square well for a particle in an arbitrary  state.

r

I II

0

Then the radial equation becomes:

The solution is:

II.  The Hydrogen Atom

Plug these in: then the Radial Equation becomes

Again the solution is RII(ρII)= Cj(ρII)+ Dn(ρII)

Since Region II does not include r=0, we do no thave to discard the Neumann for n

Since we have not yet fixed C and D, we can rewrite RII as a different form of linear combination of j and n

−−12

II.  The Hydrogen Atom

Ignore the phases for now

Recall eiθ = cosθ + isinθ

For a specific V0, a, u, and , one can solve this for the bound states Ei

I The Hydrogen Atom

What this means is "the eigen functions and eigen energies

possible for the electron in a one-electron atom"

This is just the central potential problem again, now for

V= −Ze2

r

So we know that Ψelectron will have the form Ψ  R(r)Ym

because of this (, m), also subscript the ′ Ψ :Ψm

So the most general Ψ must be a linear combination of all possible Ψm

energy levels

To find R, recall the Radial Equation (Form 2):

(i) consider the case where ρ → ∞

Recall a physically acceptable Ψ

II.  Facts about the principal quantum number n

⎝⎜ ⎞⎠⎟( + 1)ρH − e

− 2

⎛

⎝⎜ ⎞⎠⎟ρ+1H

and divide out e

− 2

*Notice that for a given value of , this is an eigenvalue equation for H with eigenvalue = λ

Recall H here is part of a wavefunction and not a hamiltonian

(n + 1)(2 + n + 2) a n Recursive realtion for the ai

This relationship between the ai is like the one for an exponential function

To see this compare:

The "H(ρ)" series "The eρ series", eρ = ρi

To force u to be a physically acceptable wavefunction, truncate the series H

34


Facts about λ=n

1 "n" is called the Principal Quantum Number

2 Recall the definition

4 Notice no matter how large n is, its En will always be slightly < 0, so at lease weakly bound

So this potential V=−Ze2

r has an infinite number of bound energy states (unlike the square well).

5 Energy degeneracies:

(i) due to  → The energy depends only on n, but for each n, there are n possible values of 

 = 0, 1, , n-1

(ii) due to m→ For each , m can be -, -+1, , 0, , -1, 

(iii) total due to m and  is then

6 A (2+1)-fold degeneracy of m-values is characteristic of a spherically symmertric potential

Recall we found that

*It is common to derive a0 ≡ −2

µe2, the Bohr radius

Read 2 handouts

Read Chapter 14

Facts about the Ψe in hydrogen :

(i) They are totally orthogonal:

∫Ψn′ ′′mΨn m dVolume= δn n′ δ ′ δm m′

due to the eimϕ in the Ym

due to the orthogonality of Legendre Polynomials P

recall Ym=(Normalization constant)⋅ eimϕ ⋅ Pm

The Laguerre Polynomials are orthogonal in n

(ii) Recall what a complete set of  functions is: it can act as a basis for its space′

i.e., any possible wavefunction in that space can be written

as a linear combination of the elements of the basis

The comlete set of eigen functions of the hydrogen atom look like:

So the bound states do not form a complete set by themselves

(iii) Notice all the Rn  r

So for >0, Rn(r = 0) = 0

No probability of finding the e− at the origin in these states

for  = 0, Rn(r= 0) = constant

So the ground state e− has a spherical probability distribution which includes the

origin So the ground state e− has finite probability to be found inside the nucleus.

(iv) To calculate the probability of finding the e− at a specific r, calculate

Probability(r)=r2 | Rn |2

This is similar to P(x)=|Ψ(x)|2

The r2 adjusts to spherical coordinates

E>0 (scattering) states (we will study these in Chapter 23)

E<0 (bound) states

V = −Ze

r2

When you calculate r2 | R |2 you find

(v) to calculate the probability of finding the e− at a particular θ,

calculate Probability( θ)=|Ym

( θ,ϕ) |2

sin θdθdϕ

since the ϕ appears in eimϕ it

will disappear from the probability

Review Syllabus

Read Goswami section 13.3 and chapter 14 plus the preceding chapter as necessary for reference

Read Chapter 14

II.  The effect of an EM field on the eigen functions and eigen values of a charged particle

III.  The Hamiltonian for the combined system of a charged particle in a general EM field

I Probability current for an e− in Hydrogen

Recall the definition of probability current:

Recall this describes the spatial flow of probability

→ NOT necessarily the motion of the point of maximum probability

→ definitely not the motion of a particle whose probability of location is related to Ψ

Calculate J for the e− in Hydrogen:

To show this recall from EM:

Consider a differentially small piece of it which is located at a distance  r from the origin.

This piece forms an element of a current loop which is at least momentarily circulating relative to the origin This is physically identical to a magnetic dipole whose magnetic dipole moment m is given by: 

where Je

circulates

∫  Je( 

r )dVolume

we can convert our probability current 

J into a physical charged current 

Je by multiplying by the charge:

II.  The Hamiltonian for the combined system of a charged particle in a general EM field

III.  The Hamiltonian for a charged particle in a uniform, static B field

Read Chapter 15

we will find that H  p

2

2 µ + [stuff]⋅ B+[stuf ′f]B

2

we will study each kind separately

II The Hamiltonian for the combined system of a charged particle in a general EM field

II.  The Hamiltonian for a charged particle in a uniform, static B field

III.  The normal Zeeman effect

(iii) Plug in L to get H, then convert everything possible to operators

Carry this out:

To find L, recall that usually

L=T-V

Here T= 1

2 µ v2 (use µ for mass everywhere)

What is v? Recall the EM field has 2 kinds of potential:

scalar potential ϕ and vector potential 

A How to combine them? (we can't just say "V= ϕ + 

F=q  E+ q

∂A i

∂t +

q

v c

∂A

∂x i −

q c



v⋅∇Ai

This equation concerns component i (i=1, 2, or 3)

of a vector equation Generalize to the full vector equation

[A,p]= i∇ ⋅ A

now recall that for a static 

B field, 

∇ ⋅A= 0 Show this:

Recall in general B can be produced by (i) a uniform current and (ii) a ∂E

∂t .Consider the static case, so A=f(I) only

Recall Stoke's Theorem:

for any vector v,

∂x etc not

∂

∂ ′x

I The Hamiltonian for an e− in a uniform, static B field (continued)

II The normal Zeeman effect

III Response to the e− to the B2 term

IV Summary of e− response to static uniform Bˆz

V The discovery of spin

call it 

B=|B|ˆz

constant, 1 directional, no position dependence

In general for any B,

Recall the vector identity:

u⋅(y ⋅ z) = (u ⋅ y) ⋅ z

"µ" is the reduced mass of the system

*If you want the answer in mks units, set c=1

I The normal Zeeman effect

II Response to the e− to the B2 term

III Summary of e− response to static uniform Bˆz

IV The discovery of spin

We will study the effect of each term separately upon the e's wavefunction and energy

II The Normal Zeeman Effect

Recall He in uniform static B in ˆz = p2

B field

Recall each energy level "n" is degenerate,

all of its  and m levels have the same energy

How does this H1 affect the e's energy levels?

mΨn m quantum number "m" NOT mass

call this collection

of constants "ωL",

the Larmor frequency

So the presence of H1 means that the different "m" levels are no longer degenerate;

each has its own energy given by:

2  2n2 m=0

E of all levels −µz 2e4

2  2n2 − ωL m=-1 with same n m=-2

etc.

The fact that a magnetic field can cause the levels designated by "m" change energy causes "m" to be

called "The magnetic quantum number"

II Response of the e- to the ~B 2 term

this is identical to the Harmonic Oscillator:

III Summary of e response to static uniform Bˆz

So far we found that

[L z , H2D

HO]= 0

So all the terms of H have the same e function

call it "Ψnmk"

Plug it in:

Set = 0 for now

H2D, HO

I The Discovery of Spin

II Filtering particles with a Stern-Gerlach apparatus

III Experiments with filtered atoms

III The discovery of spin

Suppose you wanted to measure the total angular momentum of a particle

call it 

J as in Ch 11 (Note: this J is not a current)

We showed in Chapter 13 that angular momentum ∝ magnetic moment

−e L

2µc =



M

Now call this " 

J" to be general, i.e to allow for more than just orbital angular momentum

M is in a magnetic field 

B it feels a force on itwhich depends on the relative orientation of 

Now if a particle with moment Mz is in the apparatus, it will feel a force ∝ Mz

If the particle is moving through the apparatus, the force will deflect its path from a straight line

and its deflection will measure Mz

deflection ∝ M z ˆz

F ∝ M z

particle

apparatus

An apparatus like this is called a Stern-Gerlach Experiment

If you accumulate a large number of identical particles, you can find out what

∝ the apparent orientation of the object

Mz = mcosθ

particle's magnetic dipole

Stern and Gerlach mad a beam of silver atoms

=0,

m=0

M z

M

I The Discovery of Spin

II Filtering particles with a Stern-Gerlach apparatus

III Experiments with filtered atoms

Read handout from Feynman lectures and Chapter 15

spot 1

spot 2

smear that didn't appear

M cannot have arbitrary orientation: otherwise instead of 2 spots there would have been a continuous

smear reflecting that all possible Mz states were present

2 each outer e− had a non-zero Mz which was not related to its orbital angular momentum

that had been arranged to have =m=0

call this "non-orbital angular momentum" = spin Its quantum number is s

and its "orbital angular momentum" m = ms

3 Recall for orbital angular momentum, the possible values m can take are -,  + 1, , 0, ,  − 1, 

so the number of possible values of m is (2 + 1)

4 here, experimentally it was found that the number of possible ms values is 2

Recall that regular angular momentum is quantized in the direction its allowed to have, so that

I Filtering particles with a Stern-Gerlach apparatus

if you pass them through a Stern-Gerlach device, they will separate into

different beams, one beam for each value.

Notice if you obstruct all but 1 output path, you can produce a beam that is

pure m z = +1

Example:

pure m z = +1

When a beam is put into a definite state like this,

it (the output of the SG) is called a "prepared beam"

or a "filtered beam" or a "polarized beam"

Now make a slightly modified SG that can return the polarized beam

to the original axis of travel

modified SG device

this barrier is moveable so we could select any mz value

this shows what is blocked

S this gives a particular device a name in case more than 1 is in series

Make up symbols for the prepared states:

what come out of S

is the final state we are checking to see IF they have label final states with bras

This one is < -1|

Other possibilities for this system are < 0 | or < +1|

*Note S can represent both the device that prepares a particle's state, or the state itself